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Question-103338




Question Number 103338 by I want to learn more last updated on 14/Jul/20
Commented by som(math1967) last updated on 14/Jul/20
OC=14−r  again OC=r(√2)  ∴r(√2)=14−r   r((√2)+1)=14  ∴r=14((√2)−1)cm  Area of shaded part  =(1/4)×π×14^2 −π×14^2 ((√2)−1)^2 cm^2
$$\mathrm{OC}=\mathrm{14}−\mathrm{r} \\ $$$$\mathrm{again}\:\mathrm{OC}=\mathrm{r}\sqrt{\mathrm{2}} \\ $$$$\therefore\mathrm{r}\sqrt{\mathrm{2}}=\mathrm{14}−\mathrm{r} \\ $$$$\:\mathrm{r}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)=\mathrm{14} \\ $$$$\therefore\mathrm{r}=\mathrm{14}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{cm} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\pi×\mathrm{14}^{\mathrm{2}} −\pi×\mathrm{14}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{cm}^{\mathrm{2}} \\ $$
Commented by I want to learn more last updated on 14/Jul/20
Wow, thanks sir, i appreciate.
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$
Commented by harckinwunmy last updated on 14/Jul/20
are you from Nigeria?
$$\mathrm{are}\:\mathrm{you}\:\mathrm{from}\:\mathrm{Nigeria}? \\ $$
Commented by Study last updated on 14/Jul/20
why OC=r(√2)
$${why}\:{OC}={r}\sqrt{\mathrm{2}} \\ $$
Commented by som(math1967) last updated on 14/Jul/20
Commented by som(math1967) last updated on 14/Jul/20
OBCA is a square with side r  ∴OC=r(√2)
$$\mathrm{OBCA}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{side}\:\mathrm{r} \\ $$$$\therefore\mathrm{OC}=\mathrm{r}\sqrt{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 15/Sep/21
nice
$$\mathrm{nice} \\ $$

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